Question

The centre of a circle passing through the point $(0,0),(1,0)$ and touching the circle $x^2+y^2=9$ is ?

• A. $(\frac{3}{2},\frac{1}{2})$
• B. $(\frac{1}{2},\frac{3}{2})$
• C. $(\frac{1}{2},\frac{1}{2})$
• D. None of these

Answer 1

The correct option is D None of these

Let the circle be$x^2+y^2+2gx+2fy+c=0$

$\because$ It passes through $(0,0)$

$\Rightarrow 0+c=0$

$\Rightarrow c=0$

It also passes through $(1,0)$

$\Rightarrow 1+0+2g+0+c=0$

$\Rightarrow g=\frac{-c-1}{2}=\frac{-1}{2}$

$\therefore$Circle $\to x^2+y^2-x+2fy=0$

Centre$\to (\frac{1}{2},-f)$

Other circle : $x^2+y^2=9$

Centre \rightarrow (0,0)$

Radius $=3$

$\because Circles touches internally.

$\therefore$ Difference between radius = Distance between centres.

Radius of smaller circle $=\sqrt{g^2+f^2-0}=\sqrt{\frac{1}{4}+f^2}$

$\Rightarrow 3-\sqrt{\frac{1}{4}+f^2}=\sqrt{(-g-0{)}^2+(-f-0{)}^2}$

$\Rightarrow 3-\sqrt{\frac{1}{4}+f^2}=\sqrt{\frac{1}{4}+f^2}$

$\Rightarrow 2\sqrt{\frac{1}{4}+f^2}=3$

$\Rightarrow \frac{1}{4}+f^2=\frac{9}{4}$

$\Rightarrow f^2=\frac{8}{4}=2$

$\Rightarrow f=\pm \sqrt{2}$

Hence centre $\to (-g,-f)$

$\Rightarrow (\frac{1}{2},\pm \sqrt{2})$

$\therefore$None of the options is correct.